Does zero times a discontinuous function equate to another discontinuous function?

Question

For f(x): R -> R, where f(x) is a discontinuous function.

As example let f(x) = e^x / x,when x = 0, f(0) = g(0), with g(x) being a continuous function only defined at x = 0, then:

0*f(x) = 0*(e^x / x) = 0, for x not equal to 0

and 0*f(0) = 0*g(x) =0, for when x = 0

then 0*f(x) = 0 for all x (which becomes a continuous function)

And this will always be true for all discontinuous functions.

Would any of my reasoning be wrong?

Answer 1

One cannot write $f:{\bf{R}}\rightarrow{\bf{R}}$ by just $f(x)=e^{x}/x$ without specifying the value of $f$ at $x=0$, that is, then how does $f(0)$ defined in this case? Indeed, you must specify a number to $f(0)$, say, $f(0)=c$, then the zero function multiplies with $f$ is simply $x\rightarrow 0\cdot e^{x}/x=0$ for $x\ne 0$, and $0\rightarrow 0\cdot c=0$ for $x=0$.

Of course, one need no to define a function $f$ with domain ${\bf{R}}$ each time. We can talk about, say, $g:{\bf{R}}-\{0\}\rightarrow 0$, $g(x)=e^{x}/x$, then the resulting multiplication $0\cdot g$ is defined on ${\bf{R}}-\{0\}$ with the rule that $x\rightarrow 0\cdot e^{x}/x=0$ for $x\in{\bf{R}}-\{0\}$. In this case, it makes no sense to ask what is the value of $0$ times $g(0)$ because the domain of $g$ has ruled out the discussion about the value of $g$ at $x=0$.

Answer 2

Answer to the edited question:

I think your reasoning is correct. Your way of writing it is a little strange.

If $g(x)$ is only defined at $x = 0$, then you are introducing unnecessary notation. Just write "$f(0) = (\text{whatever value goes here})$". I might write "Let $f(x) = \mathrm{e}^x/x$, for $x \neq 0$ and $f(0) = a$ for any real number $a$. Then $f$ is continuous for all $x \neq 0$ and is discontinuous at $x = 0$. Then $0 \cdot f = 0$ is the function $0$, continuous for all $x$." Note that the function $0$ is the function that receives some number of arguments, ignores them, and returns zero.

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Previous answer:

Your difficulty with $f(x) = \frac{\mathrm{e}^x}{x}$ is not that it is discontinuous; it is continuous on its entire domain. Your difficulty is that the domain of $f$ is not all real numbers; it's $\mathbb{R} \smallsetminus \{0\}$.

The product $0 \cdot f(x)$ is only defined if $0$ and $f(x)$ are defined, but $f$ is only defined on its domain. So the product is only defined on $\mathbb{R} \smallsetminus \{0\}$.

Consequently, $0 \cdot f(x) = 0$ for any $x$ in $\mathbb{R} \smallsetminus \{0\}$, the domain of $f$, and is undefined everywhere else.

However, here is another discontinuous function: $$ g(x) = \begin{cases}0 &, x < 0 \\ 1 & , 0 \leq x \end{cases} \text{.} $$ Notice that the domain of $g$ is all real numbers, $\mathbb{R}$. Also, $g$ is discontinuous at $x = 0$ -- its value jumps discontinuously from $0$ to $1$. So $0 \cdot g(x)$ is defined for all $x$ and is $0$.

Summary : a function can be discontinuous without having a point "missing" from its domain. The product $0 \cdot f(x)$ is defined everywhere $f$ is, so on the domain of $f$. Just because $f$ is discontinuous at $x = x_0$ does not mean $x_0$ is omitted from the domain of $f$. Thus, $0 \cdot f(x)$ can be defined and is $0$ at discontinuous points of $f$ in the domain of $f$.